1. If one wishes to accumulate a Php 30,000-fund in 5 years, how much should he deposit now at 18% compounded quarterly?

2. If a bank offers a rate of 4.5% compounded semiannually, how much should you deposit to accumulate Php 50,000 in 15 years

3. Suppose that a particular radioactive substance has a half-life of 25 years. Starting with 100 grams of the substance, how many grams of the substance would remain after 15 years?

To calculate the answers to these questions, we will be using the formula for compound interest and exponential decay.

1. To find out how much should be deposited now to accumulate a Php 30,000-fund in 5 years with 18% interest compounded quarterly, we can use the formula for compound interest:

A = P * (1 + r/n)^(n*t)

Where:
A = the future value of the fund after time t
P = the principal amount (initial deposit)
r = annual interest rate (in decimal form)
n = number of times the interest is compounded per year
t = number of years

In this case, we have:
A = Php 30,000
r = 18% or 0.18 (expressed as a decimal)
n = 4 (quarterly compounding)
t = 5 years

Let P be the amount to be deposited now. We need to solve for P.

Php 30,000 = P * (1 + 0.18/4)^(4*5)

Simplifying the equation:

30,000 = P * (1.045)^20

Dividing both sides by (1.045)^20:

P = 30,000 / (1.045)^20
P ≈ Php 13,197.72

So, approximately Php 13,197.72 should be deposited now to accumulate a Php 30,000-fund in 5 years at 18% compounded quarterly.

2. To find out how much should be deposited to accumulate Php 50,000 in 15 years with a 4.5% interest rate compounded semiannually, we can use the same formula:

A = P * (1 + r/n)^(n*t)

Where:
A = the future value of the fund after time t
P = the principal amount (initial deposit)
r = annual interest rate (in decimal form)
n = number of times the interest is compounded per year
t = number of years

In this case, we have:
A = Php 50,000
r = 4.5% or 0.045 (expressed as a decimal)
n = 2 (semiannual compounding)
t = 15 years

Let P be the amount to be deposited now. We need to solve for P.

Php 50,000 = P * (1 + 0.045/2)^(2*15)

Simplifying the equation:

50,000 = P * (1.0225)^30

Dividing both sides by (1.0225)^30:

P = 50,000 / (1.0225)^30
P ≈ Php 22,161.05

So, approximately Php 22,161.05 should be deposited to accumulate Php 50,000 in 15 years at a 4.5% interest rate compounded semiannually.

3. To find out how many grams of the radioactive substance would remain after 15 years, we can use the formula for exponential decay:

N = N0 * (1/2)^(t/h)

Where:
N = the remaining amount of the substance after time t
N0 = the initial amount of the substance
t = number of years
h = the half-life of the substance

In this case, we have:
N0 = 100 grams
t = 15 years
h = 25 years

We need to solve for N.

N = 100 * (1/2)^(15/25)

Simplifying the equation:

N = 100 * (1/2)^(3/5)

N ≈ 100 * 0.69314

N ≈ 69.314 grams

So, approximately 69.314 grams of the substance would remain after 15 years with a half-life of 25 years.